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Simplify x^2 - 9 / 2x^2 - 3x -9

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Sorry I'm trying to write this out in proper equation
\[\frac{ x^2 - 9}{2x^2 - 3x -9}\]
Like that right?

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Yes that's it
Couldn't dot he divide thing
\[\frac{(x-3)(x+3)}{(x-3) (2 x+3)}\] \[\frac{\cancel{(x-3)}(x+3)}{\cancel{(x-3)} (2 x+3)}\]
Could you explain like step by step, sorry I'm not very good at this
What u have to do is, u have to factor the both numerator and denominator..
After factoring, u will get, \[\frac{(x-3)(x+3)}{(x-3) (2 x+3)}\] Now u have to cancel the like terms, leaving u with, \[\frac{x+3}{2x+3}\]
Thankyou I get it now, as soon as you factorised it's easy from there thanks a lot!
No problem! ^_^

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